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Oct 28, 2024
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feat: update solutions to lc problem: No.0687 #3682

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179 changes: 86 additions & 93 deletions solution/0600-0699/0687.Longest Univalue Path/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -60,9 +60,19 @@ tags:

### 方法一：DFS

相似题目：
我们设计一个函数 $\textit{dfs}(root)$，表示以 $\textit{root}$ 节点作为路径的其中一个端点，向下延伸的最长同值路径长度。

- [543. 二叉树的直径](https://github.com/doocs/leetcode/blob/main/solution/0500-0599/0543.Diameter%20of%20Binary%20Tree/README.md)
在 $\textit{dfs}(root)$ 中，我们首先递归调用 $\textit{dfs}(root.\textit{left})$ 和 $\textit{dfs}(root.\textit{right})$，得到两个返回值 $\textit{l}$ 和 $\textit{r}$。这两个返回值分别代表了以 $\textit{root}$ 节点的左孩子和右孩子为路径的其中一个端点，向下延伸的最长同值路径长度。

如果 $\textit{root}$ 存在左孩子且 $\textit{root}.\textit{val} = \textit{root}.\textit{left}.\textit{val}$，那么在 $\textit{root}$ 的左孩子为路径的其中一个端点，向下延伸的最长同值路径长度应为 $\textit{l} + 1$；否则，这个长度为 $0$。如果 $\textit{root}$ 存在右孩子且 $\textit{root}.\textit{val} = \textit{root}.\textit{right}.\textit{val}$，那么在 $\textit{root}$ 的右孩子为路径的其中一个端点，向下延伸的最长同值路径长度应为 $\textit{r} + 1$；否则，这个长度为 $0$。

在递归调用完左右孩子之后，我们更新答案为 $\max(\textit{ans}, \textit{l} + \textit{r})$，即以 $\textit{root}$ 为端点的路径经过 $\textit{root}$ 的最长同值路径长度。

最后，$\textit{dfs}(root)$ 函数返回以 $\textit{root}$ 为端点的向下延伸的最长同值路径长度，即 $\max(\textit{l}, \textit{r})$。

在主函数中，我们调用 $\textit{dfs}(root)$，即可得到答案。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点个数。

<!-- tabs:start -->

Expand All @@ -76,16 +86,16 @@ tags:
# self.left = left
# self.right = right
class Solution:
def longestUnivaluePath(self, root: TreeNode) -> int:
def dfs(root):
def longestUnivaluePath(self, root: Optional[TreeNode]) -> int:
def dfs(root: Optional[TreeNode]) -> int:
if root is None:
return 0
left, right = dfs(root.left), dfs(root.right)
left = left + 1 if root.left and root.left.val == root.val else 0
right = right + 1 if root.right and root.right.val == root.val else 0
l, r = dfs(root.left), dfs(root.right)
l = l + 1 if root.left and root.left.val == root.val else 0
r = r + 1 if root.right and root.right.val == root.val else 0
nonlocal ans
ans = max(ans, left + right)
return max(left, right)
ans = max(ans, l + r)
return max(l, r)

ans = 0
dfs(root)
Expand Down Expand Up @@ -114,7 +124,6 @@ class Solution {
private int ans;

public int longestUnivaluePath(TreeNode root) {
ans = 0;
dfs(root);
return ans;
}
Expand All @@ -123,12 +132,12 @@ class Solution {
if (root == null) {
return 0;
}
int left = dfs(root.left);
int right = dfs(root.right);
left = root.left != null && root.left.val == root.val ? left + 1 : 0;
right = root.right != null && root.right.val == root.val ? right + 1 : 0;
ans = Math.max(ans, left + right);
return Math.max(left, right);
int l = dfs(root.left);
int r = dfs(root.right);
l = root.left != null && root.left.val == root.val ? l + 1 : 0;
r = root.right != null && root.right.val == root.val ? r + 1 : 0;
ans = Math.max(ans, l + r);
return Math.max(l, r);
}
}
```
Expand All @@ -149,22 +158,22 @@ class Solution {
*/
class Solution {
public:
int ans;

int longestUnivaluePath(TreeNode* root) {
ans = 0;
dfs(root);
int ans = 0;
auto dfs = [&](auto&& dfs, TreeNode* root) -> int {
if (!root) {
return 0;
}
int l = dfs(dfs, root->left);
int r = dfs(dfs, root->right);
l = root->left && root->left->val == root->val ? l + 1 : 0;
r = root->right && root->right->val == root->val ? r + 1 : 0;
ans = max(ans, l + r);
return max(l, r);
};
dfs(dfs, root);
return ans;
}

int dfs(TreeNode* root) {
if (!root) return 0;
int left = dfs(root->left), right = dfs(root->right);
left = root->left && root->left->val == root->val ? left + 1 : 0;
right = root->right && root->right->val == root->val ? right + 1 : 0;
ans = max(ans, left + right);
return max(left, right);
}
};
```

Expand All @@ -179,29 +188,28 @@ public:
* Right *TreeNode
* }
*/
func longestUnivaluePath(root *TreeNode) int {
ans := 0
var dfs func(root *TreeNode) int
func longestUnivaluePath(root *TreeNode) (ans int) {
var dfs func(*TreeNode) int
dfs = func(root *TreeNode) int {
if root == nil {
return 0
}
left, right := dfs(root.Left), dfs(root.Right)
l, r := dfs(root.Left), dfs(root.Right)
if root.Left != nil && root.Left.Val == root.Val {
left++
l++
} else {
left = 0
l = 0
}
if root.Right != nil && root.Right.Val == root.Val {
right++
r++
} else {
right = 0
r = 0
}
ans = max(ans, left+right)
return max(left, right)
ans = max(ans, l+r)
return max(l, r)
}
dfs(root)
return ans
return
}
```

Expand All @@ -223,28 +231,19 @@ func longestUnivaluePath(root *TreeNode) int {
*/

function longestUnivaluePath(root: TreeNode | null): number {
if (root == null) {
return 0;
}

let res = 0;
const dfs = (root: TreeNode | null, target: number) => {
if (root == null) {
let ans: number = 0;
const dfs = (root: TreeNode | null): number => {
if (!root) {
return 0;
}

const { val, left, right } = root;

let l = dfs(left, val);
let r = dfs(right, val);
res = Math.max(res, l + r);
if (val === target) {
return Math.max(l, r) + 1;
}
return 0;
let [l, r] = [dfs(root.left), dfs(root.right)];
l = root.left && root.left.val === root.val ? l + 1 : 0;
r = root.right && root.right.val === root.val ? r + 1 : 0;
ans = Math.max(ans, l + r);
return Math.max(l, r);
};
dfs(root, root.val);
return res;
dfs(root);
return ans;
}
```

Expand All @@ -271,19 +270,23 @@ function longestUnivaluePath(root: TreeNode | null): number {
// }
use std::cell::RefCell;
use std::rc::Rc;

impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, target: i32, res: &mut i32) -> i32 {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, target: i32, ans: &mut i32) -> i32 {
if root.is_none() {
return 0;
}

let root = root.as_ref().unwrap().as_ref().borrow();
let left = Self::dfs(&root.left, root.val, res);
let right = Self::dfs(&root.right, root.val, res);
*res = (*res).max(left + right);
let root = root.as_ref().unwrap().borrow();
let left = Self::dfs(&root.left, root.val, ans);
let right = Self::dfs(&root.right, root.val, ans);

*ans = (*ans).max(left + right);

if root.val == target {
return left.max(right) + 1;
}

0
}

Expand All @@ -292,13 +295,9 @@ impl Solution {
return 0;
}

let mut res = 0;
Self::dfs(
&root,
root.as_ref().unwrap().as_ref().borrow().val,
&mut res,
);
res
let mut ans = 0;
Self::dfs(&root, root.as_ref().unwrap().borrow().val, &mut ans);
ans
}
}
```
Expand All @@ -320,16 +319,15 @@ impl Solution {
*/
var longestUnivaluePath = function (root) {
let ans = 0;
let dfs = function (root) {
const dfs = root => {
if (!root) {
return 0;
}
let left = dfs(root.left),
right = dfs(root.right);
left = root.left?.val == root.val ? left + 1 : 0;
right = root.right?.val == root.val ? right + 1 : 0;
ans = Math.max(ans, left + right);
return Math.max(left, right);
let [l, r] = [dfs(root.left), dfs(root.right)];
l = root.left && root.left.val === root.val ? l + 1 : 0;
r = root.right && root.right.val === root.val ? r + 1 : 0;
ans = Math.max(ans, l + r);
return Math.max(l, r);
};
dfs(root);
return ans;
Expand All @@ -347,29 +345,24 @@ var longestUnivaluePath = function (root) {
* struct TreeNode *right;
* };
*/

#define max(a, b) (((a) > (b)) ? (a) : (b))

int dfs(struct TreeNode* root, int target, int* res) {
int dfs(struct TreeNode* root, int* ans) {
if (!root) {
return 0;
}
int left = dfs(root->left, root->val, res);
int right = dfs(root->right, root->val, res);
*res = max(*res, left + right);
if (root->val == target) {
return max(left, right) + 1;
}
return 0;
int l = dfs(root->left, ans);
int r = dfs(root->right, ans);
l = root->left && root->left->val == root->val ? l + 1 : 0;
r = root->right && root->right->val == root->val ? r + 1 : 0;
*ans = max(*ans, l + r);
return max(l, r);
}

int longestUnivaluePath(struct TreeNode* root) {
if (!root) {
return 0;
}
int res = 0;
dfs(root, root->val, &res);
return res;
int ans = 0;
dfs(root, &ans);
return ans;
}
```

Expand Down
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